Implicit plane curve discretization

I want to have a matrix with 2 lines and n columns with the coordinates of points that lie on the implicit curve where is a function. I know that the matlab function fimplicit can plot the curve. So in its subroutines it grabs a set of points from\near that curve. How can I extract those points and use them for future calculations (like the perimeter of the curve)?

Thanks!

Answer 1

[Matt J]

fp=fimplicit(@(x,y) x.^2 + 2*y.^2 - 1); %Example

Points=[fp.XData;fp.YData]

Points = 2×587

-0.0458 -0.0533 -0.0667 -0.0800 -0.0933 -0.1067 -0.1200 -0.1333 -0.1467 -0.1600 -0.1689 -0.1733 -0.1867 -0.2000 -0.2133 -0.2267 -0.2337 -0.2400 -0.2533 -0.2667 -0.2800 -0.2834 -0.2933 -0.3067 -0.3200 -0.3251 -0.3333 -0.3467 -0.3600 -0.3616 -0.7063 -0.7061 -0.7055 -0.7048 -0.7040 -0.7031 -0.7020 -0.7008 -0.6994 -0.6980 -0.6969 -0.6964 -0.6947 -0.6928 -0.6908 -0.6887 -0.6875 -0.6864 -0.6840 -0.6815 -0.6788 -0.6781 -0.6760 -0.6730 -0.6699 -0.6687 -0.6667 -0.6632 -0.6597 -0.6593

[Maxim Bogdan]

Thanks for the good answer!

But is it possible to get only the points and not the graphic?

[Walter Roberson]

fp = fimplicit(@(x,y) x.^2 + 2*y.^2 - 1, 'visible', 'off'); %Example
Points = [fp.XData;fp.YData]
delete(fp)

However this creates a graphic object -- so for example if you have no existing figure then one will be created with an axes in it (that just will not have any visible content.)

[Maxim Bogdan]

That works fine for me!

It doesn't open any figure that code for me.

Thanks a lot!